Calculating in closed form an integral in Airy function Can we hope for a nice closed form for the integral below?
$$\int_0^1 \frac{\displaystyle \text{Ai}\left(-\frac{t}{2^{2/3} \sqrt[3]{3-3 t}}\right)^2+\text{Bi}\left(-\frac{t}{2^{2/3} \sqrt[3]{3-3 t}}\right)^2}{ \sqrt[6]{1- t}} \, dt$$
What tools do you recommend me to use? 
Some information about Airy function here https://en.wikipedia.org/wiki/Airy_function.
EDIT: let me know if anything is wrong with the question, or if it happens that Airy function is annoying to some extent. I'm glad to improve the post. When I have ideas about the way to go,
I let you know. What else?
 A: This is only a partial solution. I haven't obtained any closed form, but reduced to an integral of an elementary function.
One can simplify the above integral as follows
$$
I=\int_0^1 \frac{\displaystyle \text{Ai}\left(-\frac{t}{\sqrt[3]{12(1 -t)}}\right)^2+\text{Bi}\left(-\frac{t}{ \sqrt[3]{12(1- t)}}\right)^2}{ \sqrt[6]{1- t}} \, dt.
$$
Using the integral representation (see the book Vallee, Soares "Airy function and applications to physics")
$$
\operatorname{Ai}^2(t)+\operatorname{Bi}^2(t)=\frac{1}{\pi^{3/2}}\int_0^\infty\frac{1}{\sqrt{x}}e^{xt-\frac{x^3}{12}}dx
$$
one gets:
$$
I=\frac{1}{\pi^{3/2}}\int_0^1 \frac{dt}{ \sqrt[6]{1- t}} \int_0^\infty\frac{1}{\sqrt{x}}\exp\left\{ -\frac{xt}{ \sqrt[3]{12(1- t)}}-\frac{x^3}{12}\right\}dx=\\
\frac{12^{1/6}}{\pi^{3/2}}\int_0^1 dt \int_0^\infty\frac{1}{\sqrt{x}}e^{ -xt-x^3(1-t)}dx=\\
\frac{12^{1/6}}{\pi^{3/2}}\int_0^\infty\frac{dx}{\sqrt{x}}e^{-x^3}\int_0^1 e^{ (x^3-x)t}dt =\\
\frac{12^{1/6}}{\pi^{3/2}}\int_0^\infty\frac{e^{-x}-e^{-x^3}}{x^3-x}\frac{dx}{\sqrt{x}}
$$
A: Assuming that we know the Taylor series of $\text{Ai}^2(x)$ and $\text{Bi}^2(x)$ in a neighbourhood of the origin, the problem boils down to evaluate a weigthed sum of values of the Beta function. Since the Airy functions are the fundamental solutions of:
$$ y'' = x y $$
if $y=\text{Ai}$ or $y=\text{Bi}$ we have:
$$ \frac{d}{dx}y^2 = 2yy',\quad \frac{d^2}{dx^2}y^2 = 2y'^2+2x yy',\quad \frac{d^3}{dx^3}y^2=(2x+1)2yy'+2xy'^2+2x^2y^2$$
so $\text{Ai}^2$ and $\text{Bi}^2$ fulfill the third-order ODE:
$$ y''' = 2x^2 y+(1+2x-x^2)y'+x y'' $$
that can be used to find the coefficients of the Taylor series.
Or maybe there is some slick way to exploit the Parseval's identity, once known that the Fourier transform of $\text{Ai}(x)$ is $e^{-\frac{is^3}{3}}$.
